PDE II Homework 2 Solutions

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1 PDE II Homewor 2 Solutios Exercise 1: Order ad wea topologies Let u ad v be two sequeces of L p fuctios, covergig to u ad v i wea-l p. ssume that for ay, u v. Prove that u v. Solutio: Note that if u ad v coverge to u, v i wea L p, the sice for ay of fiite measure, we have 1 L q. It follows that udx = lim u dx lim v dx = vdx. Therefore, ad so u v a.e. (v u)dx, for every of fiite measure, Exercise 2: Uiform covexity of L 2. Let u be a bouded sequece i L 2 (R) wealy covergig to u. ssume as well that u L 2 coverges to u L Decompose u u 2 L 2 usig the ier product o L 2 ad coclude that u u strogly. 2. Give a example of a sequece u covergig wealy to u i L 1, s.t. u L 1 coverges to u L 1, but u does ot coverge strogly to u. Solutio: 1. Wea covergece i L 2 meas (u, u ) u 2 L 2. Therefore, u u 2 L 2 = u 2 L 2 + u 2 L 2 2(u, u ).. 2. Choose f = 1 [,π] i L 1 ad let f = (1 + si(x))f. Note that f f ad f 1 f 1, by the Riema-Lebesgue lemma. However, f does ot coverge strogly to f, sice f f 2 2 = π si 2 (x)dx = π/2. Exercise 3: Semicotiuity Defie for ay > 1. F (u) = u(x) φ(x)dx, for some φ L (R) with φ. 1

2 1. Show that F is covex. 2. Show that F is cotiuous i the L orm. 3. Show that F is lower semi-cotiuous for ay L p orm. Solutio: 1. Covexity of F follows from the uiform covexity of. 2. We will use the followig elemetary iequality for > 1, x y x y ( x 1 + y 1 ) x, y R. Suppose that u u i L, the we fid ( u u )φdx φ u u ( u 1 + u 1 )dx. If gives = 1, a applicatio of Hölder s iequality ad the fact that u 1 = u 1 u u ( u 1 + u 1 )dx u u ( u 1 + u 1 ). Therefore sice u 1 + u 1 lim is uiformly bouded i we may coclude that u φdx = u φdx. 3. Let u u i L p, ote that up to extractio of a subsequece we may assume that lim F (u ) = lim if F (u ). However, u u i L p implies that there exists a subsequece u such that u u poitwise a.e. Sice φ, Fatou s lemma implies, F (u) lim if F (u ) = lim F (u ) = lim if F (u ). Hece F is lower semi-cotiuous. 4. By part 3 ad part 1 we ow that the sub-level sets of F, {F α} are covex ad strogly closed i L p. Mazur s Theorem (a trivial applicatio of hyperplae separatio) the implies that {F α} is also wealy closed. Therefore F is wealy lower semicotiuous. Exercise 4. Uiform Covexity agai. Let > 1, u be a wealy covergig sequece to u i L p (I) with p >, I compact. s u is uiformly bouded i L p/, we may assume that it is wealy covergig to U. 1. Prove that U u. Hit: Use the the previous exercise. 2. ssume that i additio U = u. Prove that for ay 1 l the the wea limit of 2

3 u l is u l. Hit: Show that U (w lim if u l ) /l. 3. ssume agai that U = u ad that N, eve. Prove that the wea limit of (u u) is. Coclude that u coverges strogly i L. 4. (More difficult). Prove the same thig without assumig N. 5. Show that if u coverges wealy to u i L p with 1 < p < + ad if u L p coverges to u L p, the u coverges strogly. Solutio: 1. By usig the previous exercise, we ow that for ay positive, cotiuous Φ Φ ( u U) dx. Therefore U u. 2. Extract ay subsequece s.t. u σ() l is wealy covergig i L p/l to some quatity U l. By the previous questio U l u l. O the other had u σ() = ( u σ() l ) /l with /l 1 so still by the previous questio U U /l l. So U U /l l u = U which proves that all quatities are equal s is eve the 2 so by questio 2, u 2 coverges to u 2 wealy i L p/2 (I) ad as I is compact i L 1 (I). Hece u u 2 dx = ( u 2 + u 2 2u u) dx. Therefore u coverges to u i L 2. Now by Hölder estimates u u L u u θ L 2 u u 1 θ L p with θ/2 + (1 θ)/p = 1/ ad hece θ >. This cocludes that u coverges to u i L First of all ote that if 2 the the previous questio still wors. The difficulty here is 1 < < 2. For simplicity let us assume that u here. first possibility as setched i the correctio give i class is to use questio 2: We ow that u l coverges wealy to u l for ay 1 l. The oe uses the desity of the vector defied by {v l, 1 l } amog cotiuous fuctios less tha v ear ad ot icreasig too fast at ifiity. Let us istead use this opportuity to preset a differet ad slightly more formal approach but oe related to the importat cocept of Youg measures. Defie for ay a χ(v, a) = I v a. Up to a extractio, deote χ(x, a) a wea limit of χ(u (x), a). 3

4 Remar that χ(v, a) = a ψ(v, a) with ψ(v, a) = v a if a v ad ψ(v, a) = if a > v. Now for ay a >, ψ is covex ad cotiuous i v. Deote ψ(x, a) a wea limit of ψ(u (x), a), oe has ψ(u(x), a) ψ(x, a), ψ = a χ. O the other had sice > 1 Therefore ψ(u(x), a) a 2 da = v = ( 1) = u ( 1) = ψ(v, a) a 2 da. U ( 1) = lim ψ(x, a) a 2 da. ψ(u (x), a) a 2 da This implies that the wea limit of ψ(u (x), a) is ψ(u(x), a) for almost every a ad hece that χ is equal to χ(u(x), a). Now observe that χ 2 = χ ad compute χ(u (x), a) χ(u(x), a) 2 dx da = χ(u (x), a) dx da + χ(u(x), a) dx da 2 χ(u (x), a) χ(u(x), a) dx da. Taig the limit, we deduce that χ(u (x), a) coverges strogly to χ(u(x), a) i every L q space. To coclude ote that u (x) u(x) dx u (x) u(x) 1 χ(u (x, a)) χ(u(x), a) dx da. 5. Deote by U the wea-* limit i measures of u p. Oe has still u p U. Moreover u p L = u p (x) p dx U(dx) = u p L = u(x) p dx. p Hece U = u p. By the previous questios u coverges to u i ay L q with q < p. Now deote α(l) = sup u (x) p I u L dx If α(l) does ot coverge to as L teds to the the wea-* limit of u p caot belog to L 1 ad therefore oe would have U u p. Cosequetly α(l) as L. Sice for ay fixed L u u p I u L dx, the oe cocludes that u coverges to u i L p. 4

5 Exercise 5. No-separability of M 1 ([, 1]) 1. ssume we have µ M 1 ([, 1]), a [, 1] such that µ δ a = µ δ a ([, 1]) < 1. Show that there exists c > s.t µ = cδ a + ν with ν δ a. 2. ssume that µ is a dese sequece i M 1 ([, 1]). Costruct a oe to oe applicatio φ : [, 1] N ad deduce a cotradictio. Solutio: 1. Note that sice µ δ a < 1, we have that µ({a}). lso if ν = µ µ({a})δ a, the ν({a}) = ad so ν δ a. Therefore µ = µ({a})δ a + ν. 2. Suppose that there exists a dese subset {µ } M 1 ([, 1]). For each a [, 1] let φ(a) N be such that µ φ(a) δ a < 1/2. To see that the φ : [, 1] N is oe-to-oe, suppose there are a, b [, 1], a b, such that φ(a) = φ(b). We fid that 2 = δ a δ b δ a µ φ(a) + δ b µ φ(b) < 1 which is a cotradictio. Therefore φ is a oe-to-oe mappig of [, 1] to N, which is also a cotradictio sice [, 1]ad N do t have the same cardiality. Therefore M 1 ([, 1]) is o-separable. 5